Question

How do you find parametric equations for the path of a particle that moves along the `x^2 + (y-3)^2 = 16` Halfway around counterclockwise, starting at (0,7). 0 ≤ t ≤ pi?

Answer 1

`(x, y) = (-4 sin T, 3 + 4 cos T)`, so that, when T = 0,

the position is (0, 7).,

Explanation:

The parametric equations for the circle

#(x-a)^2+(y-b)^2=R^2 are

x = a + R cos t and

y= b + R sin t.

Here, they are

x = 4 cos t and

y= 3 + 4 sin t.

At `(0, 7), t = pi/2.`

Referring to this point as initial point, the parametric equations are

obtained by the substitution `T = t-pi/2`.

The transformed equations are

`x=4cos(T+pi/2)=--4 sin t and`

#y = 3 + 4 sin (T+pi/2)=3+4cos T.